Equation for accurate prediction of PCR yield

It is a cliche of freshman biology labs to point out that "every cycle of PCR doubles the DNA, so the yield will be $2^{cycles}$ times the template amount". However, if this were true, 1 ng of template would generate about 35 billion ng after 35 cycles, or 35 grams of DNA. This is clearly absurd and not the case.

Of course, the power-of-2 claim is a gross oversimplification (if anything, it is an upper bound - but even so, a very uninformative one), and in practice, yields will fall far short of it because:

• Every single duplex of DNA does not denature at each cycle
• Primers do not bind to every single molecule of DNA at each cycle
• Not every DNA strand gets bound by a polymerase at every cycle
• Not every polymerase that binds manages to complete the entire product in time in every cycle
• The reaction inhibits itself by depleting dNTPs
• The heat denatures the reaction by degrading enzyme

In fact, cursory examination of qPCR output often follows saturation kinetics: Mathematical methods for modeling qPCR are obviously well developed.

My question is about ordinary PCR: Is it possible get a reasonable expectation of nanogram yield for an ordinary PCR done in a tabletop cycler, with typical PCR reagents?

For instance, when amplifying from a plasmid, I would like to calculate how many cycles to do, how much template to use, and how much product to load on the agarose gel to ensure that I will be able to clearly distinguish exponential amplification (both primers anneal), linear amplification (only one primer anneals), and no amplification (neither primer can anneal or the reaction did not work).

An expected efficiency for a typical PCR is 80%, meaning each cycle multiplies the copy number of the targeted DNA sequence 1.58 times.

Firstly, it makes more sense to refer to the amount of DNA in a polymerase chain reaction in terms of copy number or in terms of moles; the number of DNA molecules of interest is what the reaction is operating on, and the mass of product generated is a function of the length of the product (and, to a lesser degree, on the composition of the product).

The following discussion is sourced from this URL: https://www.csun.edu/~hcbio027/biotechnology/lec3/pcr/p.htm

According to Perkin-Elemer, copy-number amplification of 100,000 fold of the targeted sequence of DNA can be expected from a PCR with 0.1 ng of Lambda phage DNA (a well-characterized and standard DNA isolate) in a 100 µL reaction with > 25 cycles of denaturation, annealing, and extension.

In the above 100,000-fold amplification example, if the targeted amplicon were to be 500 bp in length, the estimated molecular-weight of duplex DNA of 500bp is 325,000 g/mol (based an average base-pair having a molecular mass of 650 g/mol).

The Lamdba Phage genome is 42,502 base-pairs in length. 42,502 bp × 650 grams/mol/bp = 2.762×10^7 grams/mol.

0.1 ng Lambda DNA -> 0.1×10^-9 grams. 0.1×10^-9 g ÷ 2.762×10^7 g/mol = 3.619 × 10^-18 moles. 3.619 × 10^-18 moles × NA (Avogradro's Number) = 2.179×10^6 copies, or 2,179,000 copies.

2.179×10^6 copies × 100,000 = 2.179×10^11 copies. 2.179×10^11 copies ÷ NA × 325,000 g/mol = 1.176×10^-7 grams of sequence. 1.176×10^-7 grams is equal to 0.117 µg or 117 ng.

An amplification yield of 100,000x after 25 cycles would mean at each cycle 1 template would yield 1.58 templates for the next round of synthesis.

How was this calculated? If c is the number of copies made per round of synthesis, then:

c^25 = 100,000 = 10^5
so c^5 = 10
and so 5(log c) = log 10 = 1
so log c = 0.2 and c = 1.58 (approximately)

(Or you could calculate the 25th root of 100,000 on a calculator, if you prefer.)

If we obtain 1.58 copies instead of the theoretical maximum of 2 copies, then the efficiency of the reaction could be said to be 79% (because 1.58/2.00 = 0.79).

One reason this calculation is important is that a slight loss of efficiency is magnified through the amplification. A reaction may appear to have not worked if the efficiency drops (in each cycle) by just a few percent. Optimization is critically important in the polymerase chain reaction.

• Here is a useful website for calculating the molecular mass and number of copies of double-stranded DNA molecules based on their length in base-pairs. cels.uri.edu/gsc/cndna.html – Patrick Jan 12 '16 at 20:08

The equation is correct, but there's an additional asymptotic limit to a maximum concentration of product depending on the starting concentration of NTPs, template and primer pairs in solution too.